Power Supply Resevoir Size

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Bear in mind that for audio frequencies below 100Hz (or 120Hz) the reservoir caps get topped up during an audio cycle. Worst case is actually a 50Hz (or 60Hz) square wave aligned with the mains cycle, so the cap has to provide all the voltage/current for the flat top. This assumes that the cap does indeed get topped up fully during each mains half-cycle - a really huge cap fed from a weedy transformer might not.
What exactly does this mean for the psu-design? Besides the fact that you'll need a beefy transformer?

In the amp I am planning to build, I expect some 6A current at extended full load into a 4Ohm speaker. This will probably never happen. My 500VA transformers are rated for 6.25A current, nominal. I think this means that it can supply this current for longer than just short bursts. Not sure though. I think this should be okay. What are your thoughts?
 
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The PSU has a different job. It is converting Mains AC to isolated DC with as near to constant non-varying voltage as you can afford.

After rectification the PSU has enormous ripple, going from zero volts to Vpk of the LV output.

[...]
Does this mean that the caps directly after the rectifiers need to be more durable than the ones further down the line?
 
I don't try to calculate nor use the inductance value.
I just appreciate the rounding of the ripple as seen on the scope.
That tells me that much of the Buzz is being attenuated.
I can see less ripple. One I measured yesterday with 150T gave 36mVpp of very spikey ripple at the first capacitor and 12mVpp of rounded ripple at the second capacitor.
It was actually an AAK PCB with the intercapacitor fuse location replaced with the coil.

Have you looked at average current?
Is that 6A, a DC value, or an an AC value or a peak value?
Some of the peak comes from the on board local decoupling.
Most of the remainder of the peak comes from the last capacitor in the rCRC.
The r passes only the average.
The R passes quite a bit more than average but not nearly near peak level.
If an L+r is used the L impedes the change in current even more. So it passes nearer the average current rather than towards the peak value.
 
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Inputting the numbers you gave above:
C=80mF and t=160ms results in R=2 Ohms. That equates to a 12V voltage drop at full output. That is, ofcourse, unacceptable.
Why did you jump from a proposed 0r11 to 2r?
I already gave an example for much less current than you are predicting using 0r5 to 0r8 and went on to say you would be using less to suit your 4ohms speakers.

Your 280W into 4ohms is more than double the current demand of my 100W into 8ohms.
 
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Does this mean that the caps directly after the rectifiers need to be more durable than the ones further down the line?
Very definitely. I have posted dozens of times that a two stage filter bank imposes very high ripple on the first capacitor. It must be selected to survive this duty.
A bank of 4*1mF is better than a singleton @ 4mF
Even two 2m2F would probably be better than an expensive 4m7F
 
What exactly does this mean for the psu-design? Besides the fact that you'll need a beefy transformer?

In the amp I am planning to build, I expect some 6A current at extended full load into a 4Ohm speaker. This will probably never happen. My 500VA transformers are rated for 6.25A current, nominal. I think this means that it can supply this current for longer than just short bursts. Not sure though. I think this should be okay. What are your thoughts?
280W into 4r0 is equivalent to 47.3Vpk and 11.8Apk.
6A is not the amplifier demand on it's PSU capacitors.

A reactive speaker will demand a lot more than 11.8Apk that an equivalent resistor would demand.
expect a direct driven driver to demand at least 1.5times the resistor and expect a crossovered 2way or 3way to demand at least 2times the resistor and often exceeding that. Rare events can demand 5times the resistor current and the vast bulk of that will come from the local decoupling for the few usec that it will last at the highest levels.
 
Again, I am humbly put back in school :) The 6A I mentioned, comes from I=P/U. But I suppose I have to factor in the 4Ohm resistance.

You're right in that max I may be up to two or more times higher than the number for a purely resistive load. I read that before, but didn't factor it in, here. Earlier I mentioned to expect peaks of more than 22A at full load. Since I suspected that the bulk of that peak current will be taken care of by the local decoupling (as just confirmed by AndrewT), I went on to think that the the current draw from the psu will be somewhat more civilized and more constant than the draw from the local decoupling caps. But, as usual, there's probably more to it than that.

The 2Ohms for the pi-filter comes from the RC calculation with 160 or so msec you mentioned a few posts earlier. If you think that 0.1R to 0.2R is perfectly fine, I could just do with the basic design for the UniversalPSU V3 (with pi-resistors) as can be found here on the forum, right? No need for the coil, apart from additional HF suppression. This is getting slightly confusing...
 
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I don't try to calculate nor use the inductance value. I just appreciate the rounding of the ripple as seen on the scope.
That tells me that much of the Buzz is being attenuated. I can see less ripple. One I measured yesterday with 150T gave 36mVpp of very spikey ripple at the first capacitor and 12mVpp of rounded ripple at the second capacitor. It was actually an AAK PCB with the intercapacitor fuse location replaced with the coil.

Have you looked at average current? Is that 6A, a DC value, or an an AC value or a peak value? Some of the peak comes from the on board local decoupling. Most of the remainder of the peak comes from the last capacitor in the rCRC. The r passes only the average. The R passes quite a bit more than average but not nearly near peak level. If an L+r is used, the L impedes the change in current even more. So it passes nearer the average current rather than towards the peak value.

The current demand of a power amplifier will be (I'm thinking out loud, here), a modulated DC per rail. Both the positive and negative rails, will remain positive and negative respectively. But the amount of current drawn will vary with demand, which is determined by the input signal and the load at the outputs. But, isn't it the task of the local decoupling to more or less "smoothen" the load of the PSU? The vast majority of the transient current peaks will be covered by these capacitors, right? Again, please correct me when I'm wrong.

So, let's say that the quiescent current of the amp is a bit higher than the bias of the outputs, at, say 200mA per rail. This is a DC component that will always be present (positive and negative rails respectively). The audio signal will be superimposed on this DC component.

Now, let's say that at max power transients there will indeed be a current demand of about 22A. This cannot be handled by the local decoupling caps alone. The PSU will have to assist with this. I'm theorizing that you'll need large capacitors for this *after* the coil, since you want the current through the coil to be as DC as possible to minimize reactance and voltage drop. The capacitors between the rectifiers and coil can be a bit smaller, to provide the charging current for the larger capacitors. So, I would say the the current through the coil approaches average current (although it varies a bit with demand). Again, I'm thinking out loud here. So please correct me when I'm talking crap.

[Edit] In the above, the L can be substituted with an R. Though the effect on the ripple I'll have to determine for myself, using a scope.

By the way: the ripple you say to have observed, was that with the amplifier connected to the psu?
 
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[...]

The NFB cut-off frequency should be at least half an octave below the cut-off frequency of the input filter.
The PSU cut-off frequency should be at least half an octave below the NFB cut-off frequency.

This gives me the following.
Flat passband = 20Hz to 20kHz
F-1dB passband = 4Hz to 100kHz
F-3db band is 2Hz to 200kHz

NFB filter <=2Hz/SQRT(2)
PSU filter <=NFB filter/SQRT(2).

RC of 2Hz is 79.6ms
NFB >=113ms
PSU >=160ms

Why did you jump from a proposed 0r11 to 2r?
tau = R.C --> R = tau/C

R = 0,160sec / 0,040F --> R = 4Ohms per rail (!!!).

That can't be right...

I already gave an example for much less current than you are predicting using 0r5 to 0r8 and went on to say you would be using less to suit your 4ohms speakers.

Your 280W into 4ohms is more than double the current demand of my 100W into 8ohms.
So, if I will choose R instead of L, I can do with 4 or 5 parallelled 0R47's or so?
 
to choke or not to choke?

Hello,
Lundahl does make a nice double coil choke called the LL1694 2A ( '' my dealer '' in Germany supplies two different ll1694 also one with lower current rating).
Depending on how much ac there is across the choke the current rating can be bigger. Putting the two colis in parallel you will get a 30 millihenry choke with a 0,45 serie resistance. I am sure it filters a lot better than the air coil that pops up here sometimes.
Greetings, Eduard
 
Sorry guys but,
...2110 posts and nearly 3 years later and we're STILL talking about numbers and nothing about the qualities of the actual components ....

A bank of 4*1mF is better than a singleton @ 4mF
Even two 2m2F would probably be better than an expensive 4m7F
(post #2105)

Andrew, where on earth did this come from? Maybe some context?
 
quality

Hello,
I know the lundahls are good and will be better than any air coil in that spot. Dont know what brand of caps are used. I like the panasonic tsup type.
Usually the technically advanced and good quality build will also sound the best.
And of course people should use decent transformer not just go for a big VA rating toriodal one. I like the R core from selectronic in France but they stop at 500va.
Greetings, Eduard
 
Sorry guys but,
...2110 posts and nearly 3 years later and we're STILL talking about numbers and nothing about the qualities of the actual components ....

A bank of 4*1mF is better than a singleton @ 4mF
Even two 2m2F would probably be better than an expensive 4m7F
(post #2105)

Andrew, where on earth did this come from? Maybe some context?
The cheap way to buy ripple capacity is to buy smaller capacitors and parallel them.
No clever arithmetic just add up the ripple capacity of 4off 1mF and compare to one 4m7F.
 
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that's the speaker impedance. That can be right.
You lost me. I thought I was calculating the needed R?? What does speaker impedance have to do with the pi-filter? It will affect the needed C, but what about the pi-R? Except that you want it low? Confused...

I think that is what this started with when you asked about what R to use.
So, based on the fact that I want low series R and need inductance for ripple suppression, could I use something like this in each power rail?
Wiring the coils in parallel gives me very low R and approximately 1.67mH. Would that work well?
 
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So, based on the fact that I want low series R and need inductance for ripple suppression, could I use something like this in each power rail?
Wiring the coils in parallel gives me very low R and approximately 1.67mH. Would that work well?
cored coils will saturate when passing DC.
If you have a dual coil cored inductor and run the currents in the coils in inverse phase then the core sees a net zero current and the inductance value is maintained.
If the coil currents are in phase then the core saturates because the core sees the DC currents as adding up.
 
You lost me. I thought I was calculating the needed R?? What does speaker impedance have to do with the pi-filter? It will affect the needed C, but what about the pi-R? Except that you want it low? Confused...
you have two separate systems.

system one is the last smoothing capacitor supplying current to the speaker via the amplifier. Here you have a load, the speaker impedance and a capacitor. The RC is a filter.

system two is the filtering of the mains AC to become near perfect DC supply current.
The AC is first converted to unipolar half wave pulses. This is then "stored" in a capacitor (C). But the circuit supplying the current has an inherent resistance (r) that is usually ignored. This r & C is effectively a filter rC.
One can add (cascade of two filters) an extra filter stage using a separate R and a separate C giving a rCRC cascaded filter. This R is filtering the mains ripple, just as the r is also filtering the mains ripple.

if I relabel the various filters to become PSU consisting of rC'RC'' and the speaker (Z) fed from C''

we have rC', RC'' and ZC''
these are all different filters.
 
choke or not?

Hello,
If you use a common mode choke you should use it as a common mode one to prevent the core from saturating and ending up with no inductance at all.
If you want an effective filter you need a choke with a big core and an air gap.
My supplier in Germany has the ll1694 2A from Lundahl It will give you a 30mH choke with 0,45 ohm serie resistance. There are some other ones with higher inductance but also higher serie resistance.
Lundahl Plate chokes, Lundhal Anodendrossel
Dont say ooh ooh 0,45 ohm there goes my power full lower notes.
Greetings, Eduard
 
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